Editing Ultraparallel theorem (section)

==Hilbert's construction==

Let {{mvar|r}} and {{mvar|s}} be two ultraparallel lines.

From any two distinct points {{mvar|A}} and {{mvar|C}} on s draw {{mvar|AB}} and {{mvar|CB'}} perpendicular to {{mvar|r}} with {{mvar|B}} and {{mvar|B'}} on {{mvar|r}}.

If it happens that AB = CB', then the desired common perpendicular joins the midpoints of AC and BB' (by the symmetry of the [[Saccheri quadrilateral]] ACB'B).

If not, we may suppose AB < CB' without loss of generality. Let E be a point on the line s on the opposite side of A from C. Take A' on CB' so that A'B' = AB. Through A' draw a line s' (A'E') on the side closer to E, so that the angle B'A'E' is the same as angle BAE. Then s' meets s in an ordinary point D'. Construct a point D on ray AE so that AD = A'D'.

Then D' ≠ D. They are the same distance from r and both lie on s. So the perpendicular bisector of D'D (a segment of s) is also perpendicular to r.<ref>{{cite book|last1=H. S. M. Coxeter|author1-link=H. S. M. Coxeter|title=Non-euclidean Geometry|date=17 September 1998|isbn=978-0-88385-522-5|pages=190–192}}</ref>

(If r and s were asymptotically parallel rather than ultraparallel, this construction would fail because s' would not meet s. Rather s' would be limiting parallel to both s and r.)